Author Topic: CHALLENGE : Rectangles  (Read 21677 times)

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Kerry

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Re: CHALLENGE : Rectangles
« Reply #75 on: September 29, 2006, 09:31:40 PM »
seant,

Just had a quick look at your PDF ... I believe your method fails at axiom 3 ..

C^2 + D^2 = W^2  is NOT necessarily correct.

.. unless I misunderstand your calculations.
kdub, kdub_nz in other timelines.
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SEANT

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Re: CHALLENGE : Rectangles
« Reply #76 on: September 29, 2006, 10:16:18 PM »
You understand the calculation but the scope of this formula may be more limited than you're expecting.  At this point I'm strictly looking at a "peg" with the shortest possible hypotenuse.  I'm assuming the length of this short hypotenuse is equal to the Width of the "hole".  The calculations and spreadsheet only deal with that particular peg.

I was hoping that the information from that rectangle and the "hole" rectangle would alllow some form of direct interpolation to a target peg size.  In fact, if the two ranges 0 to 258.81905, and 1000 to 1118.034 are assigned to the sine wave 0 to Pi/2, other peg lengths can be derived; but with a regular error pattern, peaking at 2%.  The next step would be to account for that error.

I'll post that sine overlay tomorrow.

Devoting this much thought to a problem of dubious practical value may not be healthy, however.  :ugly:

 
« Last Edit: September 29, 2006, 10:18:05 PM by SEANT »
Sean Tessier
AutoCAD 2016 Mechanical

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Re: CHALLENGE : Rectangles
« Reply #77 on: September 30, 2006, 12:55:45 PM »

I can create a relationship but it's a relationship of unknowns so i can't derive an answer.


By using the fact that the triangles are similar, I was able to reduce the equations to a single unknown.

Assume Angle A is the larger angle in the triangles, h = hole height (shorter length), w = hole width (longer length), d = peg depth (short side of peg).  Then the length L (long side of the peg) could be determined by:

L = (h - dsinA)/cosA

It could also be determined by:

L = (w - dcosA)/sinA

The problem is determining the value of A.  Combining the above equations, I get messes like:

tanA = (w - dcosA) / (h - dsinA)

...which I sure can't figure out how to solve for A, except numerically with a computer for any given w, h, and d.

SEANT

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Re: CHALLENGE : Rectangles
« Reply #78 on: October 02, 2006, 05:34:20 AM »
Ah yes,

I can see where those equations came from.  But, the claim that only one unknown remains is a bit misleading.  Both magnitude of L and angle A are undetermined.

I also see how the final equation was derived, by setting the first two equal to themselves. Unfortunately I think that only gives you the definition of tangent, i.e. Sin/Cos of any angle.  Though, I may be wrong there.

What would make life easy is if a second, generaly unrelated equation, could be formulated - to fulfill the "2 equations, 2 Unknowns" requirement.  By analyzing the "Peg" rectangle hypotenuse I get:

L = ((H-(2D*SIN(A))2 + W2 - D2)1/2

The major problem is setting it equal to one of your equation.  The reduction process becomes quite a handful. 

 
« Last Edit: October 02, 2006, 05:37:03 AM by SEANT »
Sean Tessier
AutoCAD 2016 Mechanical

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Re: CHALLENGE : Rectangles
« Reply #79 on: October 02, 2006, 10:47:59 PM »

What would make life easy is if a second, generaly unrelated equation, could be formulated - to fulfill the "2 equations, 2 Unknowns" requirement.


But, that's exactly what I did.  Here are the two equations, with two unknowns (L and A):

L = (h - dsinA)/cosA
L = (w - dcosA)/sinA

The problem is that there are two "A" terms on the right side in each equation.  That's the part I'm not sure how to handle.  Normally, to solve two equations with two unknowns, we would simply subtract them.  Unfortunately, subtracting the above two equations creates this mess:

(h - dsinA)/cosA = (w - dcosA)/sinA

We know that "A" must be less than 90 degrees, so that could simplify the problem (we can ignore any solutions in other quadrants).  But even taking that into account, I have no idea how to solve that mess for "A".  However, it is very possible to use a computer to keep trying values for A until the left and right sides of the equation are within a specific tolerance of each other.  That gives the value of A, which can be used to find L.

SEANT

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Re: CHALLENGE : Rectangles
« Reply #80 on: October 03, 2006, 07:07:05 AM »
Actually, the reason I considered them related was due to my attempt to solve them with the definitions of Sin and Cos:

Sin(A) = Opposite/Hypotenuse
Cos(A) = Adjacent/Hypotenuse

Using that method, with the drawing below, returned something as useful as CosA=CosA.



For that matter, I don't think the "unrelated" equation of my previous post is actually all that unrelated.  More likely, it is just an inefficient method of describing the general relationships you provided.

It is proving quite difficult to isolate either Sin or Cos to some rational value.
« Last Edit: October 03, 2006, 07:13:23 AM by SEANT »
Sean Tessier
AutoCAD 2016 Mechanical